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I am trying to solve the book "Linear Algebra " written by S H Friedberg, where it is given:

Prove $AB$ is invertible implies both $A $ and $B $ are both invertible ($A $ and $B $ both are square matrices).

My try----$AB $ is invertible so $\det (AB) $ is not $0$ implies $\det (A) $ and $\det (B)$ both are nonzero. So both $A $ and $B $ are invertible. But that book did not say the idea of determinant of matrix before that problem.So I believe we can prove that result not using determinant, but how?

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In the finite-dimensional case, here's another proof to add to the mix. The product $AB$ is equal to the matrix

$$[Ab_1~|~Ab_2~|\dots|~Ab_n]$$

where the $b_i$ are the columns of $B$. Since $AB$ is invertible, it follows that these columns are linearly independent. That is, no linear combination of $Ab_i$ equals the zero vector. Therefore,

$$c_1(Ab_1) + \dots + c_n(Ab_n) = A(c_1b_1 + \dots + c_nb_n)$$

will never be zero, implying that the columns of $B$ are linearly independent. Moreover, they form a basis. Consequently, $A$ must also be invertible, otherwise some choice of $c_i$ could send the sum to zero which cannot happen.

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If $AB$ is invertible, then there exists a $C$ such that $(AB)C=I$. By associativity, this is the same as $A(BC)=I$ ...

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Proof that $B$ is invertible:

If $B$ is not invertible, there exists some $x\neq 0$ so that $Bx=0$. Therefore, $(AB)x=A(Bx)=0$ and $AB$ is not invertible. Contradiction.

Proof that $A$ is invertible is almost the same (now that you know $B$ is invertible)

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  • $\begingroup$ then $A=B^(-1) $and B is invertible implies $B^(-1) $ is also invertible, and $A $ is invertible $\endgroup$ – Subhajit Saha Jul 31 '17 at 13:29
  • $\begingroup$ @SubhajitSaha $A=B^{-1}$? Why do you think that? $\endgroup$ – 5xum Jul 31 '17 at 13:30
  • $\begingroup$ You have proved $B $ is invertible,then I have done it $\endgroup$ – Subhajit Saha Jul 31 '17 at 13:45
  • $\begingroup$ @SubhajitSaha $B$ is invertible, sure, but $A=B^{-1}$ is not true... $\endgroup$ – 5xum Jul 31 '17 at 13:46
  • $\begingroup$ How!Please explain $\endgroup$ – Subhajit Saha Jul 31 '17 at 13:48
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Consider the linear maps $u$ and $v$ of $K^n$ associated with $A$ and $B$ respectively. $AB$ is associated with $u\circ v$.

Now $AB$ invertible means $u\circ v$ is an automorphism of $K^n$, so $u$ is surjective and $v$ is injective. As we're in finite dimension, this means $u$ and $v$ are automorphisms, so $A$ and $B$ are invertible.

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$AB$ is invertible, thus there exists a matrix $C$ such that $$(AB)C=C(AB)=I$$ Thus using associativity, $$A(BC)=(CA)B=I$$ These equalities give that $A$ and $B$ are invertible.

Note, here we use that for two square matrices $AB=I \implies BA=I.$

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Let $A=A_{n\times n},B=B_{n\times n}$. We know Rank$(AB)\leq$ $\min\{$Rank$(A)$,Rank$(B)\}$.

Now Rank$(AB)=n$. Then, $\min\{$Rank$(A)$,Rank$(B)\}=n$. W.l.o.g. I am taking $\min\{$Rank$(A)$,Rank$(B)\}=$Rank$(A)$. Then Rank$(A)=n$. Then $A$ is full rank.

Now $AB$ is invertable$\Rightarrow A^{-1}AB$ is invertable$\Rightarrow B$ is invertable$\space\space\space\space\space\space\blacksquare$

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Prove this:

Suppose that $AB = I$, where $A,B$ are two matrices. Then, $BA=I$.

Suppose $ABC = I$. Then $A(BC) = I$, so $A$ is invertible. Since we also have $BCA = I$, we get $B(CA) = I$. So $B$ is also invertible.

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